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A question I just came across :
A bijection $f:X\to Y$ is a homeomorphism if $f$ and $f^{-1}$ are continuous .
Show that the map $f:[0,1]\to [a,b]$ $$f(x)=(1-x)a+xb$$ is a homomorphism...

I don't know how to go with solving to show $f^{-1}$ is continuous..

coool
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    I think you mean homEomorphism. Maybe you should first find what $f^{-1}$ is. Also, $f$ is a function of a single variable, so I think you mean $f(x)=(1-x)a+xb$. – Hayden Oct 15 '14 at 18:11
  • @Hayden yes...I'm sorry about mistake ,I've made edit.. – coool Oct 15 '14 at 18:17

3 Answers3

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Hint: $$ f^{-1}(y) = x \iff f(x) = y \implies f^{-1}(a + x(b-a)) = x $$ now put $y = a + x(b-a)$ to figure out how $f^{-1}$ maps $x$.

DanZimm
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Simply compute the inverse:

$$y=x(b-a)+a\to x=\frac{y-a}{b-a}$$

that is,

$$f^{-1}(x)=\frac{x-a}{b-a}$$

that is clearly a continuous function.

ajotatxe
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In this case you can find explicit formula for $f^{-1}$:

$$y=(1-x)a+xb=a+x(b-a)$$

So for $a \neq b$:

$$x=\frac{y-a}{b-a}$$

You can check that $f^{-1}(x)=\frac{x-a}{b-a}$ (for example by checking $f(f^{-1}(x))=f^{-1}(f(x))=x$).

agha
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